We are given the equation for the rate constant K as:
\(K = \frac{2.303}{10} \log \frac{0.04}{0.03}\)
Now, simplifying the logarithmic expression:
\(K = \frac{2.303}{10} \log \frac{4}{3}\)
Using the logarithmic value:
\(K = \frac{2.303 \times 0.123}{10}\)
Now, calculate K:
\(K = 0.0285\)
The half-life t1/2 is calculated using the formula:
\(t_{1/2} = \frac{0.693}{K}\)
Substitute the value of K into the equation:
\(t_{1/2} = \frac{0.693}{0.0285} = 24.1 \, \text{s}\)
Thus, the half-life is 24.1 seconds.
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?

Chemical kinetics is the description of the rate of a chemical reaction. This is the rate at which the reactants are transformed into products. This may take place by abiotic or by biological systems, such as microbial metabolism.
The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in the concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R → P
Read More: Chemical Kinetics MCQ